Question

$\lim_{x\to\infty} \frac{2x^2+x-5}{4x^3+5x+1}$

Answer 1

0

Answer 2

To evaluate the limit of a rational function as $x$ approaches infinity, we can divide both the numerator and the denominator by the highest power of $x$ present in the denominator.

The given limit is: $\lim_{x\to\infty} \frac{2x^2+x-5}{4x^3+5x+1}$

The highest power of $x$ in the denominator is $x^3$. Divide every term in the numerator and the denominator by $x^3$: $\lim_{x\to\infty} \frac{\frac{2x^2}{x^3}+\frac{x}{x^3}-\frac{5}{x^3}}{\frac{4x^3}{x^3}+\frac{5x}{x^3}+\frac{1}{x^3}}$

Simplify the terms: $\lim_{x\to\infty} \frac{\frac{2}{x}+\frac{1}{x^2}-\frac{5}{x^3}}{4+\frac{5}{x^2}+\frac{1}{x^3}}$

As $x \to \infty$, any term of the form $\frac{c}{x^n}$ where $c$ is a constant and $n > 0$ approaches 0. Therefore, as $x \to \infty$: $\frac{2}{x} \to 0$ $\frac{1}{x^2} \to 0$ $\frac{5}{x^3} \to 0$ $\frac{5}{x^2} \to 0$ $\frac{1}{x^3} \to 0$

Substitute these limiting values into the expression: $\frac{0+0-0}{4+0+0} = \frac{0}{4} = 0$

Alternatively, for a rational function $\frac{P(x)}{Q(x)}$, as $x \to \infty$: If the degree of the numerator is less than the degree of the denominator, the limit is 0. In this case, the degree of the numerator ($2x^2+x-5$) is 2. The degree of the denominator ($4x^3+5x+1$) is 3. Since $2 < 3$, the limit is 0.